Boolean algebra is the theory behind digital hardware design. It is restricted to values 1 and 0 instead of from minus infinity to plus infinity in algebra of real values. It is the set of rules which are used to simplify the given logic expression without changing its functionality.
Boolean Algebra Rules and Laws
- Compliment Rule :
Complement of A = Ā or A’. So if A =1 then complement of A will be 0.
- AND operator
Consider the two input operation.
If you have n number of variables the you will get a 2n possible combination as the output. By referring above truth table we can say that both the operations should be high to give a output high.
A.A = A
A.0 = 0
A.1 = A
A.Ā = 0 (take the complement of another)
- OR operator
Consider the two input operation.
A + A = A
A + 0 = A
A + 1 = 1
A + Ā = 1
- Distributive Law
A.(B+C) = A.B +A.C
A+ (B.C) = (A+B). (A+C)
- Commutative Law
A + B = B + A
A.B = B.A
- Associative Law
A(BC) = (AB)C
A + (B + C) = (A + B) + C
De Morgan’s Law
There are two theorems in De Morgan’s Law. In the first theorem Morgan says when two inputs are “AND” and complemented. Then they are equivalent to the OR of the complement of the individual inputs.
Second theorem says when two inputs are “OR” and complemented, then they are equivalent to the AND of the complements of the individual inputs.
Boolean function is an algebraic expression consist of binary variables (0 and 1) and the logic operation symbols (+ ; .). for the given value of the binary variables, the function is either 1 or 0. As an example consider the Boolean function
F (x, y, z) = x + y’z
This function is equals to 1 when x= 1 or when y.z =0.1 and is equals to 0 otherwise. This function can be represented in a truth table.
Can be transformed into a circuit diagram composed of logic gates.
A Boolean function has a unique truth table. Also it has many algebraic forms which have equivalent logic. Manipulate Boolean function definition according to the rules of Boolean algebra, Why?
- To obtain a simpler expression
- Simpler and smaller circuits
- Low cost
Draw the circuit for F = XY + X’Z + YZ’ +X’YZ
Let’s simplify the F
F = XY + X’Z + YZ’ +X’YZ
= XY + X’Z + YZ’
= XY (1) + X’Z (1) + YZ'(1)
= XY(Z+Z’) + X’Z(Y+Y’) + YZ'(X+X’)
= XYZ + XYZ’ + X’YZ + X’Y’Z + XYZ’ + X’YZ’= XYZ + XYZ’ + X’YZ + X’Y’Z + X’YZ’
= XY(Z+Z’) + YZ(X+X’) + YZ'(X+X’) + X’Z(Y+Y’) + X’Y(Z+Z’)
= XY (1) + YZ (1) + YZ'(1) + X’Z (1) + X’Y (1)
= (XY+X’Y) + (YZ+YZ’) + X’Z
= Y(X+X’) + Y(Z+Z’) + X’Z
= Y (1) + Y (1) + X’Z
= Y + Y + X’Z
= Y + X’Z
By simplifying you can minimize the number of literals or term in expression. Again draw the circuit for the simplified version.
Here we can see that minimizing the expression allows the designer to use minimum number of components and it reduces the complexity of the circuit. Therefore, can reduce the cost of the design.
Minterms and Maxterms
Minterm: possible product terms
If 3 variables like X’Y’Z’, X’Y’Z, X’YZ, XY’Z’…. Each AND term is called a Minterms.
- N variables can be combined to form 2n
- Minterm is obtained from and AND term of the n variables
Maxterm: possible sum terms
If 3 variables like X+Y+Z, X+Y+Z’, X+Y’+Z…. Each OR term is called a Maxterm.
- N variables can be combined to form 2n Maxterms
- Maxterm is obtained from and OR term of the n variables
Each Maxterm is the complement of its corresponding Minterm and vice versa.
F1(x, y, z) = x’y’z + xy’z’ + xyz = m1 + m4 + m7
F2(x, y, z) = x’yz + xy’z + xyz’ + xyz = m3 +m5 + m6 + m7
Any Boolean function can be expressed as a sum of Minterms.
F1(x, y, z) = x’y’z + xy’z’ + xyz
Complement of F1 is,
F’1 (x, y, z) = x’y’z’ + x’yz’ + x’yz + xy’z + xyz’
Take the complement again + De Morgan’s Theorem,
F1(x, y, z) = (x + y + z) (x + y’ + z) (x + y’ + z’) (x’ + y + z’) (x’ + y’ +z)
Any Boolean Function can be expressed as a product of Maxterms.
Boolean functions expressed as a sum of Minterms or product of Maxterms are said to be in Canonical Form. Canonical form is not efficient, but sometimes useful in analysis and design. In an expression in canonical form, every variable appears in every form.
F (A, B, C, D) = A B’ C D’ +A’ B C’ D +A B C D
Canonical Form (Sum of Minterms)
Example: express the function f (x, y, z) = x +y’z as a sum of Minterms. If it misses one or more variables u, AND with u +u’
f(x, y, z) = x + y′z
= x(y + y′)(z + z′) + y′z(x + x′)
= (xy + xy′)(z + z′) + xy′z + x′y′z
= xyz + xy′z + xyz′ + xy′z′ + xy′z + x′y′z
= xyz + xy′z + xyz′ + xy′z′ + x′y′z
= m7 + m5 + m6 + m4 + m1
= m1 + m4 + m5 + m6 + m7
=∑(1, 4, 5, 6, 7)
Canonical Form (Product of Maxterms)
Example: express the function f (x, y, z) = xy +x’z as a sum of Minterms. Use the distributive law, x + yz = (x + y) (x + z). if it misses one or more variables u, OR with uu’.
Conversion between Canonical Forms
If f (x, y, z) =∑ (1; 4; 5; 6; 7),
- then f′ (x, y, z) =∑(0, 2, 3) = m0 + m2 + m3
- Apply De Morgan’s theorem to f′,
f (x, y, z) = (m0) ′(m2) ′(m3) ′ = M0M2M3= ∏ (0, 2, 3)
General Conversion Procedure
- Interchange the symbols ⅀ and ℿ and list those numbers missing from the original.
- n should be known, so that the total number of Minterms or Maxterms (2n) are known, to find missing terms.